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Unformatted text preview: 1 , c 2 , and c 3 such that c 1 u + c 2 v + c 3 w = (2, 0, 4) u = (3, 1, 2), v = (4, 0, 8), and w = (6, 1, 4) This means that we have 3 equations in 3 unknowns: Equation 1: 3c 1 + 4c 2 + 6c 3 = 2 Equation 2:c 1 c 3 = 0 Equation 3: 2c 1 – 8c 2 – 4c 3 = 4 From Equation 2: c 1 = c 3 , substituting we get: Equation 1: 3c 1 + 4 c 2 = 2 Equation 3: 2c 1 – 8c 2 = 4, or  c 1 – 4c 2 = 2. Adding the two equations, we get: 2c 1 = 4, c 1 = c 3 = 2. Subs. In Equation 1: 6 + 4c 2 = 2, 4c 2 = 4, c 2 = 1. Answer: c 1 = 2, c 2 = 1, and c 3 = 2...
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 Spring '06
 Anderson
 Addition, positive value, Initial Point

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